TCP/IP Cheatsheet v2.1 ()

hollowtabernacleNetworking and Communications

Oct 26, 2013 (3 years and 9 months ago)

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TCP/IP Cheatsheet v2.1
A Free Study Guide by Boson Software, Inc.
Warning: No warranty expressed or implied.
Copyright (c) 1999-2000 by Boson Software, Inc.
All Rights Reserved. Use at your own risk!
Section 1:Address Class Summary & Reserved address space
Section 2:Binary Breakdown & Decimal conversion
Section 3:Subnet Masking
Section 4:Logical Addressing
Section 5:Class C example breakdown
Section 6:Powers of 2 Shortcut
Section 7:Real World Walkthrough
Section 8:Flash Cards
Check out one of our other great Study Guides, "IP Summary Addressing Cheatsheet"
("IP-SUMMARY.PDF") for detail information on VLSM and CIDR.
Copyright (c) 2000 Boson Software, Inc. All Rights Reserved.
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Page 0
Section 1:
Address Class Summary
Range of
Number of
Number of Host
Network ID's
Internet
Networks
Per Network
(First Octet)
Class A 126 16,777,214 1 -- 126
Class B 16,384 65,534 128 -- 191
Class C 2,097,152 254 192 -- 223
The First Octet Rule: High order bits determine the Class of Address
The Most Significant (highest) bits for each Class are frozen, leaving the remainder for the Hosts portion
An Octet is a binary number of 8 bits. The smallest number is '00000000' and the largest is '11111111'
A shortcut is: 2
8
This 2 to the power of 8 is the largest octet (all 8 bits are 1's). It's 255 in decimal.
Binary Binary First Most Special
Start End Octet Sig Bits Note
Class A = 00000000 - 01111111 = Class A = 1-126 0 Assignable
Class B = 10000000 - 10111111 = Class B = 128-191 10 Assignable
Class C = 11000000 - 11011111 = Class C = 192-223 110 Assignable
Class D = 11100000 - 11101111 = Class D = 224-239 1110 Multicast
Class E = 11110000 - 11110111 = Class E = 240-247 11110 InterNIC
Total # Networks per Class
Network
Class A 0111
1111.00000000.00000000.00000000 = 2 to the 7th power -2 = 126 Net.H.H.H
Class B 1011
1111.11111111.00000000.00000000 = 2 to the 14th power = 16384 Net.Net.H.H
Class C 1101
1111.11111111.11111111.00000000 = 2 to the 21st power = 2097152 Net.Net.Net.H
Total # Hosts per Network
Host
Class A 00000000.11111111.11111111.11111111
= 2 to the 24th power - 2 = 16777214 x.255.255.255
Class B 00000000.00000000.11111111.11111111
= 2 to the 16th power -2 = 65534 x.y.255.255
Class C 00000000.00000000.00000000.11111111
= 2 to the 8th power - 2 = 254 x.y.z.255
Reserved Address Space
* RFC 1166 & 1918 = Private (internal use only) address space
Netblock
Special Use
Reference
10.X.X.X Private RFC 1918
127.X.X.X Loopback Diagnostics
172.(16-31).X.X Private RFC 1918
192.0.0.X Reserved JBP
192.0.1.X Backbone-Test-C RH6
192.0.2.X Internet-Test-C JBP
192.0.(3-255).X Unassigned NIC
192.1.(0-1).X Backbone Local Nets SGC
192.1.2.X Backbone Fiber Nets SGC
192.1.3.X Backbone Apollo Nets SGC
192.168.X.X Private RFC 1918
Copyright (c) 2000 Boson Software, Inc. All Rights Reserved.
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Page 1
Section 2:
Binary Breakdown - Decimal to Binary
Step1 is always drawing out this chart, from the right to the left, each time doubling the value starting at 1.
Most Significant Bit Weighted Values Least Significant Bit
128 64 32 16 8 4 2 1
X X X X X X X X
Step 2 is subtracting each Weighted Value from your Decmial value, until you reach 0.
Let's say we want to conver the decimal value 29 to binary. It goes like this:
Can 128 go into 29? No, so it's a 0. Can 64 go into 29? No, so it's a 0. You get the idea
Most Significant Bit Weighted Values Least Significant Bit
128 64 32 16 8 4 2 1
0 0 0 1 1 1 0 1
After going all the way down the Weighted Values chart, we now have our answer: 00011101
Binary Breakdown - Binary to Decimal
Step1 is always drawing out this chart, from the right to the left, each time doubling the value starting at 1.
Most Significant Bit Weighted Values Least Significant Bit
128 64 32 16 8 4 2 1
X X X X X X X X
Step 2 is adding up the binary values under the Weighted Values, until you reach your decimal value.
Let's say we want to conver the binary value 11000011 to decimal. It goes like this:
Draw a line under 128 and add 128, draw a line under 64 and add 64, draw a line under 2 and add 2, etc.
Most Significant Bit Weighted Values Least Significant Bit
128 64 32 16 8 4 2 1
1 1 0 0 0 0 1 1
After going all the way down the Weighted Values chart, we now have our answer: 128+64+2+1 = 195
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Page 2
Section 3:
Subnet Masking
This is the idea of taking the larger network block and chopping it into smaller pieces of equal size.
Routers require this logical segmentation to be able to address different logical subnets.
Clients require a default gateway IP address (the routers interface) to get off their local subnet.
The subnet mask is more important to a router than any other TCP/IP value.
The number of bits that are used in the subnet mask determine how many logical subnets you get.
There are a few rules involved in subnet masking.
1. The subnets cannot be all 0 or all 1 in the network or host portion of the address.
2. The all 0 subnet is called Subnet Zero, and sometimes can be used (not recommended).
3. The all 1 subnet is called the All 1's subnet, and can never be used (directed broadcast).
4. The incremental value is the IP host address starting point for the subnet.
5. Each incremental value represents another logical subnet.
6. Routers only care about the Net ID, and the subnet Broadcast.
7. Clients only care about local IP address on their subnet, and their subnet broadcast.
8. Each bit represents a power of 2. The easiest way to determine subnets is to use powers of 2.
Binary Decimal Binary Bits Number of Valid Host
Mask Mask Hosts Used Subnets Increments
00000000 0 <---> 00000000 0 Net ID Not a Sub
10000000 128* <---> 00000001 1 1 128
11000000 192 <---> 00000011 2 2 64
11100000 224 <---> 00000111 3 6 32
11110000 240 <---> 00001111 4 14 16
11111000 248 <---> 00011111 5 30 8
11111100 252 <---> 00111111 6 62 4
11111110 254* <---> 01111111 7 126 2 *
11111111 255* <---> 11111111 8 254 1 *
* = 7 subs valid for Class A or B in 1st octet. Class C has only 5 valid - the last 2 are binary all 1.
Copyright (c) 2000 Boson Software, Inc. All Rights Reserved.
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Page 3
Section 4:
Logical Addressing
* All SubNets can be broken down to: 7 ranges, 3 classes each (A, B, C)
Max. No.Math For Max. Host Subnet Mask Sub Bits Host Bits
Subnets Host ID per Subnet Required Available
Class A
0
2
24-2
16,777,214 255.0.0.0 0 24 Class A
1
2
23-2
Invalid 128 = Subnet 0 1 23 Not recommended
2
2
22-2
4,194,302 255.192.0.0 2 22 Range 1/7
6
2
21-2
2,097,150 255.224.0.0 3 21 Range 2/7
14
2
20-2
1,048,574 255.240.0.0 4 20 Range 3/7
30
2
19-2
524,286 255.248.0.0 5 19 Range 4/7
62
2
18-2
262,142 255.252.0.0 6 18 Range 5/7
126
2
17-2
131,070 255.254.0.0 7 17 Range 6/7
Class B
0
2
16-2
65,534 255.255.0.0 8 16 A' Range 7/7 & Start B
1
2
15-2
Invalid 128 = Subnet 0 9 15 Not recommended
2
2
14-2
16,382 255.255.192.0 10 14 Range 1/7
6
2
13-2
8,190 255.255.224.0 11 13 Range 2/7
14
2
12-2
4,094 255.255.240.0 12 12 Range 3/7
30
2
11-2
2,046 255.255.248.0 13 11 Range 4/7
62
2
10-2
1,022 255.255.252.0 14 10 Range 5/7
126
2
9-2
510 255.255.254.0 15 9 Range 6/7
Class C
0
2
8-2
254 255.255.255.0 16 8 B' Range 7/7 & Start C
1
2
7-2
Invalid 128 = Subnet 0 17 7 Not recommended
2
2
6-2
62 255.255.255.192 18 6 Range 1/5
6
2
5-2
30 255.255.255.224 19 5 Range 2/5
14
2
4-2
14 255.255.255.240 20 4 Range 3/5
30
2
3-2
6 255.255.255.248 21 3 Range 4/5
62
2
2-2
2 255.255.255.252 22 2 Range 5/5
Copyright (c) 2000 Boson Software, Inc. All Rights Reserved.
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Page 4
Section 5:
Class C example breakdown
It is only necessary to focus in on the octet that has been broken at the bit boundary.
Disregard all other octets, they will either be 255 (at the front) or 0 (at the end).
Remember, a decimal value of 255 means 8 bits of binary masking.
Using 2 bit subnet mask: Class C: 255.255.255.192 & Class B: 255.255.192.0 & Class A: 255.192.0.0
SUBNET
OCTET
RANGE
Incremental Value
DESCRIPTION
*Network (Subnet 0).00/000000* 0* 64 Net ID (goes in routing table)
(do not use) 00/000001* 1* First Host ID in Subnet 0
00/111110* 62* Last Host ID in Subnet 0
00/111111* 63* Broadcast for only Subnet 0
Subnetwork 1 of 4 01/000000 64 64 Net ID (goes in routing table)
01/000001 65 First Host ID for Sub 1
01/111110 126 Last Host ID for Sub 1
01/111111 127 Broadcast for only Subnet 1
Subnetwork 2 of 4 10/000000 128 64 Net ID (goes in routing table)
10/000001 129 First Host ID for Sub 2
10/111110 190 Last Host ID for Sub 2
10/111111 191 Broadcast for only Subnet 2
*Broadcast subnet 11/000000* 192* 64 Net ID (goes in routing table)
(do not use) 11/000001* 193* First Host ID in Sub All 1's
11/111110* 254* Last Host ID in Sub All 1's
11/111111* 255* Local Wire (All Subnets) Broadcast
Copyright (c) 2000 Boson Software, Inc. All Rights Reserved.
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Page 5
Section 5: CONTINUED
Using 3 bit subnet mask: Class C: 255.255.255.224 & Class B: 255.255.224.0 & Class A: 255.224.0.0
SUBNET
OCTET
RANGE
Incremental Value
DESCRIPTION
*Network (Subnet 0).000/00000* 0* 32 Net ID (goes in routing table)
(do not use) 000/00001* 1* First Host ID in Subnet 0
000/11110* 30* Last Host ID in Subnet 0
000/11111* 31* Broadcast for only Subnet 0
Subnetwork 1 of 8 001/00000 32 32 Net ID (goes in routing table)
001/00001 33 First Host ID for Sub 1
001/11110 62 Last Host ID for Sub 1
001/11111 63 Broadcast for only Subnet 1
Subnetwork 2 of 8 010/00000 64 32 Net ID (goes in routing table)
010/00001 65 First Host ID for Sub 2
010/11110 94 Last Host ID for Sub 2
010/11111 95 Broadcast for only Subnet 2
Subnetwork 3 of 8 011/00000 96 32 Net ID (goes in routing table)
011/00001 97 First Host ID for Sub 3
011/11110 126 Last Host ID for Sub 3
011/11111 127 Broadcast for only Subnet 3
Subnetwork 4 of 8 100/00000 128 32 Net ID (goes in routing table)
100/00001 129 First Host ID for Sub 4
100/11110 158 Last Host ID for Sub 4
100/11111 159 Broadcast for only Subnet 4
Subnetwork 5 of 8 101/00000 160 32 Net ID (goes in routing table)
101/00001 161 First Host ID for Sub 5
101/11110 190 Last Host ID for Sub 5
101/11111 191 Broadcast for only Subnet 5
Subnetwork 6 of 8 110/00000 192 32 Net ID (goes in routing table)
110/00001 193 First Host ID for Sub 6
110/11110 222 Last Host ID for Sub 6
110/11111 223 Broadcast for only Subnet 6
*Broadcast subnet 111/00000* 224* 32 Net ID (goes in routing table)
(do not use) 111/00001* 225* First Host ID in Sub All 1's
111/11110* 254* Last Host ID in Sub All 1's
111/11111* 255* Local Wire (All Subnets) Broadcast
Copyright (c) 2000 Boson Software, Inc. All Rights Reserved.
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Page 6
Section 6:
Powers of 2 Shortcut
Think of subnetting as stealing from Peter to give to Paul.
You have a maximum number of bits (determined by class) to play with, and if you overlap, it won't work.
Steal host bits moving from Left to Right to acquire more Subnets.
Steal subnet bits moving from Right to Left to acquire more Hosts per Sub.
Left-to-Right: Use Network n bits (2^n)-2=x to get x total number of SubNets per stolen Host bit.
Example:Steal 3 Host bits out of total 8 for Subnets on a Class C (see below)
Answer:This means you can have (2^3)-2 = (8-2) = 6 Subnets from this 3-bit Mask.
Note:The leading 3 bits (11100000) = 128+64+32 = 224 Decimal SubNet Mask.
Right-to-Left: Use remaining Host h bits (2^h)-2=x to get total x number of Hosts per SubNet.
Example:There are 5 Host bits remaining out of 8 after using 3 for Subnets on Class C (see above)
Answer:This means you can have (2^5)-2 = (32-2) = 30 Hosts per Subnet.
Note:The Subnet Mask answer would be either slash notation /27 or decimal 255.255.255.224
You are using a Class C address, with 3 bits of subnetting, and 5 host bits remaining.
Class A
Bits Req Subnets Subnets Hosts Per Hosts Per Slash Masks Sub
(n) (2^n-2) (Decimal) (2^(24-n)-2) (Decimal) (Notation) (Decimal) (Slice)
Total 24 bits 1 (2^1)-2 2-2=0 (2^23)-2 8,388,606/9 255.128.0.0 Subnet 0
to use for 2 (2^2)-2 4-2=2 (2^22)-2 4,194,302/10 255.192.0.0 Sub 1/7
subnetting 3 (2^3)-2 8-2=6 (2^21)-2 2,097,150/11 255.224.0.0 Sub 2/7
4 (2^4)-2 16-2=14 (2^20)-2 1,048,574/12 255.240.0.0 Sub 3/7
5 (2^5)-2 32-2=30 (2^19)-2 524,286/13 255.248.0.0 Sub 4/7
6 (2^6)-2 64-2=62 (2^18)-2 262,142/14 255.252.0.0 Sub 5/7
7 * (2^7)-2 128-2=126 (2^17)-2 131,070/15 255.254.0.0 Sub 6/7
8 * (2^8)-2 256-2=254 (2^16)-2 65,534/16 255.255.0.0 Sub 7/7
* = 7 subs valid for Class A or B in 1st octet. Class C has only 5 valid - the last 2 are binary all 1.
Class B
Bits Req Subnets Subnets Hosts Per Hosts Per Slash Masks Sub
(n) (2^n-2) (Decimal) (2^(16-n)-2) (Decimal) (Notation) (Decimal) (Slice)
Total 16 bits 1 (2^1)-2 2-2=0 (2^15)-2 32,766/17 255.255.128.0 Subnet 0
to use for 2 (2^2)-2 4-2=2 (2^14)-2 16,382/18 255.255.192.0 Sub 1/7
subnetting 3 (2^3)-2 8-2=6 (2^13)-2 8,190/19 255.255.224.0 Sub 2/7
4 (2^4)-2 16-2=14 (2^12)-2 4,094/20 255.255.240.0 Sub 3/7
5 (2^5)-2 32-2=30 (2^11)-2 2,046/21 255.255.248.0 Sub 4/7
6 (2^6)-2 64-2=62 (2^10)-2 1,022/22 255.255.252.0 Sub 5/7
7 * (2^7)-2 128-2=126 (2^9)-2 510/23 255.255.254.0 Sub 6/7
8 * (2^8)-2 256-2=254 (2^8)-2 254/24 255.255.255.0 Sub 7/7
* = 7 subs valid for Class A or B in 1st octet. Class C has only 5 valid - the last 2 are binary all 1.
Class C
Bits Req Subnets Max Subs Hosts Per Hosts Per Slash Masks Sub
(n) (2^n-2) (Decimal) (2^(8-n)-2) (Decimal) (Notation) (Decimal) (Slice)
Total 8 bits 1 (2^1)-2 2-2=0 (2^7)-2 0/25
255.255.255.128
Subnet 0
to use for 2 (2^2)-2 4-2=2 (2^6)-2 62/26
255.255.255.192
Sub 1/5
subnetting 3 (2^3)-2 8-2=6 (2^5)-2 30/27
255.255.255.224
Sub 2/5
4 (2^4)-2 16-2=14 (2^4)-2 14/28
255.255.255.240
Sub 3/5
5 (2^5)-2 32-2=30 (2^3)-2 6/29
255.255.255.248
Sub 4/5
6 (2^6)-2 64-2=62 (2^2)-2 2/30
255.255.255.252
Sub 5/5
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Page 7
Section 7:
Real World Walkthrough
Given: You have address 132.7.0.0. You need 5 equal-size SubNets with 1,500 Hosts Per Sub.
Objectives: Compute the following information, in the following order:
1. Find the number of Host bits to steal to get the required number of Subs.
2. Find the number of Hosts per Subnet you will get.
3. Find the decimal value of the new subnet mask.
4. Find the Incremental Value of the Subnets.
5. Find the First Host, Broadcast, and Last Host of each Subnet.
1.) Using powers of 2, first see how many bits you need to steal from Hosts to acquire 5 subnets.
a. First octet is 132, which is in the range of 128-191, so this is a Class B. Each octet has 8 bits.
b. Class B address have the first 2 octets (16 bits) locked in, so you can't touch those. There are 2 octets (16 bits) remaining.
c. Going from Left to Right, start gobbling up Host bits. If you hit 16, you are out of bounds, and it won't work.
d. Steal 1 bit? (2^1)-2=0, no that never works
e. Steal 2 bits? (2^2)-2=2, not enough subs yet, keep going
f. Steal 3 bits? (2^3)-2= 6, hey we found it, only need 3 bits for 6 subs (leaves only 1 sub for future expansion!!)
g. Write those first 3 bits into your Weighted Values chart.
Most Significant Bit Weighted Values Least Significant Bit
Subs 128 64 32 16 8 4 2 1 Hosts
1 1 1 0 0 0 0 0
2.) Using powers of 2, next see how many Host bits you have remaining.
a. Going from Right to Left, count the remaining bits. This will be how many Hosts per Sub you get.
b. Since we know we had 16 host bits total for our Class B, and we stole 3 bits, that leave us with 13 bits for Hosts.
c. Compute (2^13)-2= 8,190 Hosts Per Sub. Wow!!
d. Since this is so many more Hosts than we need, consider stealing extra Host bits to create extra Subs for future expansion!!
3.) Using the Weighted Values chart, find out what the decimal Subnet Mask will be.
a. Simply add 128 + 64 + 32 = 224 from the chart above.
b. Since this was a Class B and the first 2 octets are reserved, the default mask is 255.255.0.0
c. We stole the first 3 bits out of octet #3, so that is the only octet we really ever cared about.
d. The new decimal mask is 255.255.224.0, or shorthand notation /19 (8+8+3=19)
4.) Using the Decimal Mask, OR the last Network Bit's Weighted Value, find the Incremental Value.
a. Option 1: Take the new Subnet mask (the octet found in Step 3a above) and subtract from 256.
a. Option 2: Look at the Weighted Values chart and find the last bit flipped to 1 going Left to Right.
b. Either way, we now have the Incremental Value of 32.
c. This means Valid Subnetwork #1 is going to be 32, and each valid subnet will increase by 32, until the Subnet Mask is hit.
5.) Using the Incremental Value, count up and find the First Host IP, Broadcast for that Sub, and Last Host IP.
a. Start at the Incremental Value, which is 32.
b. Add up the next Incremental Value, which is 32 + 32 = 64.
c. Take one LESS than the NEXT Incremental value (64-1=63), and that is the PREVIOUS subnets Broadcast (for sub 32).
d. Add one to the current Incremental Value (32+1=33), that is your First Host (for sub 32).
e. Subtract one from the current subnets Broadcast (63-1=62), that is your Last Host (for sub 32).
f. See the chart below for details.
Notes: There are (2^3)-2=6 subnets created. The -2 is important, because you cannot use all 0 or all 1 subnets.
The special all 0 address is the network ID for that subnet, and will be used by a router in its routing table.
The special all 1 address is the network broadcast for all subnets on this wire.
Range 0 x 32 0 Subnet # 0 Subnet Zero should be considered invalid on any vendor's certification exam.
Range 1 x 32 32 Subnet #1 Increment 32 = Subnet ID used by Routing Table
First Host = 33
Last Host = 62
Subnet Broadcast = 63
Range 2 x 32 64 Subnet #2 Increment 64 = Subnet ID used by Routing Table
First Host = 65
Last Host = 94
Subnet Broadcast = 95
Range 3 x 32 96 Subnet #3 One less than 128 is 127, so that is the broadcast for the 96 subnet.
Range 4 x 32 128 Subnet #4 One less than 160 is 159, so that is the broadcast for the 128 subnet.
Range 5 x 32 160 Subnet #5 One less than 192 is 191, so that is the broadcast for the 160 subnet.
Range 6 x 32 192 Subnet #6 One less than 224 is 223, so that is the broadcast for the 192 subnet.
Range 7 x 32 224 Subnet # 7 Broadcast Reserved
Copyright (c) 2000 Boson Software, Inc. All Rights Reserved.
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Page 8
Section 8:
Flash Cards to Practice
(Print this page and cut out these boxes)
"Bit combinations per Mask"
"Valid SubNet Masks"
Decimal
Hosts per
Class Valid
192 Mask (64-2)=62 Any Class 1/7 11000000 2 bits 192 mask Any Class
224 Mask (32-2)=30 Any Class 2/7 11100000 3 bits 224 mask Any Class
240 Mask (16-2)=14 Any Class 3/7 11110000 4 bits 240 mask Any Class
248 Mask (8-2)= 6 Any Class 4/7 11111000 5 bits 248 mask Any Class
252 Mask (4-2)= 2 Any Class 5/7 11111100 6 bits 252 mask Any Class
254 Mask * (2-1)=1 Class A or B only 11111110 * 7 bits 254 mask Class A or B only
255 Mask * (1-1)=0 Class A or B only 11111111 * 8 bits 255 mask Class A or B only
Quiz yourself 1/2: "What is the Mask for xx Hosts in Class C?"Quiz yourself: "What are the 7 valid subnet mask bits?"
Quiz yourself 2/2: "How many Hosts for xx Mask in Class C?"
"Valid Subnets per Bit"
"Incremental Value of each Mask"
Mask bits
Increment
Subnets per
192 mask Inc. = 64 Any Class
2 bits 64 2 subnets 224 mask Inc. = 32 Any Class
3 bits 32 6 subnets 240 mask Inc. = 16 Any Class
4 bits 16 14 subnets 248 mask Inc. = 8 Any Class
5 bits 8 30 subnets 252 mask Inc. = 4 Any Class
6 bits 4 62 subnets 254 mask Inc. = 2 Class A or B only
7 bits 2 126 subnets 255 mask Inc. = 1 Class A or B only
8 bits 1 254 subnets
Quiz yourself: "What is the Incremental Value of each of the 7 Masks?"
Quiz yourself: "How many SubNets are in those 7 Masks?"
"Valid SubNets per Mask"
"Address Class Ranges"
192 mask 2 subnets Any Class Class A 1-126 Network ID 0
224 mask 6 subnets Any Class Class B 128-191 Network ID 10
240 mask 14 subnets Any Class Class C 192-223 Network ID 110
248 mask 30 subnets Any Class
252 mask 62 subnets Any Class Quiz yourself 1/3: "What is the range for Class A?"
254 mask *
126 subnets
Class A or B only Quiz yourself 2/3: "What is the range for Class B?"
255 mask *
254 subnets
Class A or B only Quiz yourself 3/3: "What is the range for Class C?"
Quiz yourself: "How many Subnets do you get on which masks?"
"Octet Bit Breakdown"
Binary 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = Decimal 255
Quiz yourself 1/2: "What is 255 in binary?"
Quiz yourself 2/2: "What are all 8 binary bits in Decimal?"
Most Significant Bit Weighted Values Least Significant Bit
Subs 128 64 32 16 8 4 2 1 Hosts
X X X X X X X X
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Page 9